A tennis ball of mass mt is held just above a basketball of mass mb, as shown in

Question

A tennis ball of mass mt is held just above a basketball of mass mb, as shown in the figure below. With their centers vertically aligned, both are released from rest at the same moment, so that the bottom of the basketball falls freely through a height h and strikes the floor. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down because the balls have separated a bit while falling.
(a)The two balls meet in an elastic collision. To what height does the tennis ball rebound? (Use any variable or symbol stated above along with the following as necessary: g.) ht=
(b) How do you account for the height in (a) being larger than h? Does that seem like a violation of conservation of energy?

Explanation / Answer

This problem is equivalent to having the basketball collide head-on with the tennis ball, when they have the same magnitude of velocity (but in opposite directions). The falling to ground just before the collison only serves to give both balls their velocity (same speed), and to reverse the direction of the basketball, so there is a collison. (In a real experiment, the tennis ball would collide with the basketball before the basketball attained full reverse speed. And of course the collisions would be significantly inelastic.) Therefore set up the problem as any elastic collison problem: Let 1 be basketball and 2 be tennis ball. Let up be the positive direction. Because ball 1 is heavier than ball 2, and because both have same speed, then ball1 will continue upward, at reduced speed, while ball 2 reverses direction. Before the collison the balls have the same speed, same magnitude of velocity, call it v0. Let balls 1 and 2 have velocities v1 and v2 after the collision. Conservation of momentum: m1 v0 - m2 v0 = m1 v1 + m2 v2 (Eqn 1) Conservation of kinetic energy: The coefficient (1/2) is factored out. m1 (v0)^2 + m2 (v0)^2 = m1 (v1)^2 + m2 (v2)^2 (Eqn 2) You now have 2 equations in 2 unknowns, v1 and v2. Plug in the values of the known mases, find the initial velocity v0 and plug itin, and solve the equatiolns for v1 and v2. Get v0 from the formula for velocity of body that falls a distance h: v = (2 g h )^1/2, (Eqn 3) where g = 9.80 m/s^2, and h = 1.10 m. So v0 = 4.64 m/s. The answer to a) is v0 = 4.64 m/s Next find height of bounce of ball 2 (tennis ball). For this you need the after-collision velocity of ball 2. Get this from equations 1 and 2. Rearrange Eqn 1 to get v1 = [ v0 (m1 - m2) - m2v2 ] / m1 (Eqn 4) Plug Eqn 4 into Eqn 2, rearrange to solve for v2. All the kinetic energy of ball 2 will be converted to gravitational potential energy as the ball rises to height h2. Rearrange Eqn 3 to get h2 = v2^2 / 2g (Eqn 5) The answer to b) is h2 of Eqn 5. Note that it was not necessary to find the velocity of ball 1 after the collison. If v1 is desired, it can be obtained by plugging the value of v2 into Eqn r to solve for v1. hope this helps

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